应用统计学大作业.docx

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1、中阂儿汕丸号（华东）CHINA UNIVERSITY OF PETROLEUM应用统计方法作业学 院：机电工程学院专 业：机械工程学 号：Z19040034姓 名：薛成智班 级：六班课堂编号：49任课教师：王清河日期：2019年11月30日= sj -= lyy - s = 357.9761111 - 110.4908445 = 247.4852666F =S)2F = S/(18-2-l)110.4908445/2247.4852666/15=3.348406738对于给定的8=010,查F(2,15)表得临界值入=2.70。由于F = 3.348406738人=2.70,所 以检验效果显著

2、，即回归方程有意义。(3)检验与,2中每个参数对y的影响的显著性。取统计量靖/W计算得F = 6.132752934,尸2 = 0.780282023,对给定的oc= 0.10,查F(2,15)表得临界值入= 2.70。由于F = 6.132752934 2,70F2 = 0.780282023 xf瓦=产 lxxf13.9389407811.66703=1.194729K = y- 龈=109.9362 - 1.194729 x 3.04243 = 106.3013 故所求的回归方程为y = 106.3013 + 1.194729/ 进行变量还原得回归方程夕=106.3013 + 1.194

3、729五检验假设“01： Si = 0s台=plx,y = 1.194729 X 13.93894078 = 16.65325Sj = lyy -略=21.21051 - 16.65325 = 4.557255=40.19652S$ 16.65325SL/Il = 4.557255/11对给定的a = 0.01,查F （1,11）表得临界值入= 9.65,由于F 入，检验效果显著，所 以拒绝“01，即方案（1）提供的回归方程有意义。2）方案2选取曲线回归（2）求解,令 = lgx,应用EXCEL可算得数据，列入表2,由表2得.表2曲线（2）数据预处理计算XX = lg Xyxi - xyt-y

4、(xi彳)3i一刃20.30103106.42-0.6166-3.516152.16804352930.477121108.2-0.4405-1.736150.76478258440.60206109.58-0.31557-0.356150.11238975650.69897109.5-0.21866-0.436150.09536729670.845098110-0.072530.063846-0.00463057980.90309109.93-0.01454-0.006158.94473E-05101110.490.0823750.5538460.045622977111.041393110

5、.590.1237680.6538460.080924907141.146128110.60.2285030.6638460.151690742151.176091110.90.2584660.9638460.249121536161.20412110.760.2864950.8238460.236027641181.2552731110.3376471.0638460.359204808191.278754111.20.3611281.2638460.456410766和11.929131429.174.715045411平均0.917625109.9362平方和1.19471721,210

6、5113z = J-P）2 = 1.194717i=l13G = 一刃2 = 21.21051i=l13lxly = W（xi - 7）（以 一刃=4.715045411i=l由此得说=号=lXXf说=号=lXXf4.7150454111.194717=3.946578Po=y-际=109.9362 - 3.946578 x 0.917625 = 106.3147 故所求的回归方程为y= 106.3147 + 3.9465787进行变量还原得回归方程y = 106.3147 + 3.946578 lg x检验假设Hoi： 1 = 0S篙=说l*y = 3.946578 x 4.71504541

7、1 = 18.6083S% = lyy -S% = 21.21051 - 18.6083 = 2.60221118.60832.602211/11=78.66052对给定的a = 0.01,查F （1.11）表得临界值入= 9.65,由于F 入，检验效果显著，所 以拒绝1,即方案（2）提供的回归方程有意义。3）方案3选取曲线回归（3）求解，令应用EXCEL可算得数据，列入表3,由表3得。表3曲线（3）数据预处理计算XX = 1/xy- yxt - xXyt - y）20.5106.420.34224-3.51615-1.20336796830.333333108.20.175573-1.736

8、15-0.30482205940.25109.580.09224-0.35615-0.03285157850.2109.50.04224-0.43615-0.01842307370.142857110-0.01490.063846-0.000951580.125109.93-0.03276-0.006150.000201601100.1110.49-0.057760.553846-0.031990236110.090909110.59-0.066850.653846-0.043710307140.071429110.6-0.086330.663846-0.057310886150.066667

9、110.9-0.091090.963846-0.087800103160.0625110.76-0.095260.823846-0.078479708180.055556111-0.10221.063846-0.108729964190.052632111.2-0.105131.263846-0.132866339和2.0508821429.17-2.101102119平均0.15776109.9362平方和0.21367121,2105113lx,x, = W(，L J/ = 0.213671i=l13(y = W(为 一刃之= 2L21051i=iy = 111.4875 - 9.8333

10、7/13lxly = W（i - 7）8-力=-2.101102119 i=l由此得ftPo = y- B3 故所求的回归方程为为少-2.1011021190.213671=-983337=109.9362 - (-9.83337) X 0.15776 = 111.4875进行变量还原得回归方程1y = 111.4875 -9.83337-x检验假设Hoi： Si = 0s台=讯4y = （-9.83337） X （-2.101102119） = 20.66092= Uv - S. = 21.21051 - 20.66092 = 0.549586 找 yy 回s 20.66092=诉=0.54

11、9586/11 = 413.5295对给定的a = 0.01,查F （1,11）表得临界值人= 9.65,由于F 入，检验效果显著，所 以拒绝1,即方案（3）提供的回归方程有意义。4）方案4选取曲线回归（4）求解,令，= lnx,应用EXCEL可算得数据，列入表4,由表4得。表4曲线（4）数据预处理计算13X= nxyxt - xyt-y（吟三）工一力20.693147106.42-1.41976-3.516154.99210471231.098612108.2-1.0143-1.736151.76097697841.386294109.58-0.72662-0.356150.25878697

12、651.609438109.5-0.50347-0.436150.21959131471.94591110-0.1670.063846-0.01066230282.079442109.93-0.03347-0.006150.00020596102.302585110.490.1896750.5538460.105050787112.397895110.590.2849850.6538460.186336485142.639057110.60.5261470.6638460.349280841152.70805110.90.595140.9638460.573623535162.77258911

13、0.760.6596790.8238460.543473728182.8903721110.7774621.0638460.827099636192.944439111.20.8315291.2638461.050924625和27,467831429.1710,85679328平均2.11291109.9362平方和6.33426921.210518 = W(吟一亍)2 = 6.334269i=i13/yy = (yf-y)2 = 21.21051 i=l13lxfy = 2(当一 ?)(% 一刃=10.85679328(=1由此得-lXfy 10.85679328周=产= c =L7139

14、77lx,xi 6.334269瓦=一瓦彳=109.9362 - 1.713977 x 2.11291 = 106.3147 故所求的回归方程为y=进行变量还原得回归方程y = 106.3147- 1.713977 Inx检验假设41： 0i = 0S=瓦=1.713977 X 10.85679328 = 18.6083SQ = lyy _= 21.21051 - 18.6083 = 2.60221118.6083S2=78.66052=78.66052F二 回_-5/11 - 2.602211/11对给定的a = 0.01,查F （1,11）表得临界值入= 9.65,由于F 入，检验效果显著

15、，所以拒绝为i，即方案（4）提供的回归方程有意义。5）方案5选取曲线回归（5）求解，令/ =以，应用EXCEL可算得数据，列入表5,由表5 得。表5曲线（5）数据预处理计算13Xxr = VxYx i - xyi-y（%-二）（%一刃21.259921106.42-0.81517-3.516152.86627649631.44225108.2-0.63285-1.736151.09871676841.587401109.58-0.48769-0.356150.17369402251.709976109.5-0.36512-0.436150.15924801471.912931110-0.162

16、160.063846-0.01035352782109.93-0.07509-0.006150.000462122102.154435110.490.079340.5538460.043942065112.22398110.590.1488850.6538460.097348041142.410142110.60.3350470.6638460.222419937152.466212110.90.3911170.9638460.376976831162.519842110.760.4447470.8238460.36640331182.6207411110.5456471.0638460.58

17、0483976192.668402111.20.5933071.2638460.749848514和26.976231429.176.725466571平均2.075095109.9362平方和2.60932921,21051lxlxr =- PT = 2.609329t=i13/yy = W(/一刃之= 2121051i=l13lxfy = 2(吟-亍)(% )=6.725466571i=i由此得=2.57747=2.57747%, -6.725466571P1 - lx,x, 2.609329瓦二9一一一 =109.9362 - 2.57747 X 2.075095 = 104.5877

18、故所求的回归方程为进行变量还原得回归方程检验假设Hoi： Bi = oy = 104.5877 - 2.57747/y = 104.5877 - 2.57747班S篙=瓦/7=2.57747 x 6.725466571 = 17.33469S篙=21.21051 - 17.33469 = 3.875822s 17.33469时而? = 3.875822/11 = 4219771对给定的a = 0.01,查F (1.11)表得临界值入= 9.65,由于F 入，检验效果显著，所 以拒绝即方案(5)提供的回归方程有意义。将上述5个方案进行整合，得下表6表6方案回归方程F1y = 106,3013 +

19、 1.194729G4.55725540.196522y = 106.3147 + 3.946578 Igx2.60221178,6605231 y = 111.4875 + 9.83337- X0.549586413.52954y = 106.3147+ 1.7139771nx2.60221178,660525y = 104.5877- 2.57747以3.87582249,19771经过对比，方案3的残差平方和最小，因而其回归方程最优，拟合效果最好，方案2 和方案4次之，方案5再次，方案1最差。习题5研究平炉炼钢的效率y与出钢量 M 和FeO(X2)的关系，测得数据如下：115.396.5

20、56.9101.0102.987.9101.4109.8103.4x214.214.614.914.918.213.213.520.013.0y83.578.073.091.483.482.084.080.088.0X1110.680.393.088.088.0108.989.5104.4101.9X215.312.914.716.418.115.418.313.812.2y86.581.088.681.585.781.979.189.980.6(1)建立y关于右,右的线性回归方程。(2)检验所建立方程是否有意义(a = 0.10)。(3)检验%2是否对y有显著影响a = 0.10)。(4)如

21、果有对y影响不显著的变量，将其去掉再建立一元回归方程。 解：(1)设所建立的线性回归方程为：夕=瓦+瓦+茂%2应用EXCEL,求得数据见表1表1数据预处理计算XlX2yxl埠XlX2x2yy21115.314.283.513294.09201.641637 269627.551185.7697225296.514.6789312.25213.161408.975271138.86084356.914.9733237.61222.01847.814153.71087.75329410114.991.410201222.011504 99231.41361.868353965102918.283.

22、410588.41331.241872.788581.861517.886955 56687.913.2827726.41174.241160.287207.81082.467247101.413.58410281.96182.251368.98517.61134705681098208012056.0440021%87841600640091034138810691.561691344.2909921144774410110615.386.512232.36234 091692189566 91323.457482 251180.312.9816448.09166.411035.876504

23、.31044.96561129314.788.68649216.091367.18239.81302.427849.96138816.481.57744268.961443.271721336.6664225148818.185.77744327.611592.8754161551,1773444915108.915.481.911859.21237.161677.068918.911261.266707.611689.518.379.18010.25334.891637.857079.451447.536256.8117104413.889.910899.36190.441440.72938

24、5 561240.628082.0118101912.280.610383.61148.841243.188213.14983.326496 36和1739.7273.61498.1171359.24240.0426470.99145351.822743.61125041.5由表1得18% = 2 吟- 18 X 万2 = 171359.2 X 1739.72 = 3217.205 i=l18l12 = W / 1 阳2 - 18 X = 26470.99- -x 1739.7 x 273.6 = 27.55 i=l1822 =乏的 - 18 x 石2 = 4240.04 -磊 x 273.6

25、2 = 81.32 i=l18lly = WyiXn - 18 x xj-y = 145351.8-x 1739.7 x 1498.1 = 560.40518l2y = Wy/i2 - 18 x 双歹=22743.61 -得 x 273.6 x 1498.1 = -27.51 i=l18力=W* - 18 x V = 125041.5- -x 1498.12 = 357.9761111t=l解正规方程3217.205瓦 + 27.55-2 = 560.40527.55瓦 + 81.32-2 = - 27.51273.6x= 72.119151118得瓦=0.1776,瓦=-0.39846,从而_1498.11739.7 ,仇=歹仇石-b2xi = 0.1776 x - (-0.39846) lolo得回归方程为y = 72.1191511 4- 0.17764 - 0.39846x2Sg =+ l2y = 560.405 x 0.1776 + (-27.51) x (-0.39846) = 110.4908445